Find the Taylor series for $\displaystyle f(x,y)=\log(4x-3y)$, up to and including the quadratic terms about the point $\displaystyle P=(1,1)$

Can someone check my working:

$\displaystyle f(1,1)=0$
$\displaystyle f_x=\frac{4}{4x-3y}\Rightarrow f_x(1,1)=4$
$\displaystyle f_y=\frac{-3}{4x-3y}\Rightarrow f_y(1,1)=-3$
$\displaystyle f_{xx}=\frac{-16}{(4x-3y)^2}\Rightarrow f_x(1,1)=-16$
$\displaystyle f_{yy}=\frac{-9}{(4x-3y)^2}\Rightarrow f_x(1,1)=-9$
$\displaystyle f_{xy}=\frac{12}{(4x-3y)^2}\Rightarrow f_x(1,1)=12$

$\displaystyle \ln(4x-3y)\approx 4(x-1)-3(y-1)+\frac{1}{2!}(-16(x-1)^2+24(x-1)(y-1)-9(y-1)^2)$

2) Using the constant and linear terms of the Taylor series above, find a first order approximation to $\displaystyle f(P+(0.01,-0.01))$

So for this would I just use

$\displaystyle \ln(4x-3y)\approx 4(x-1)-3(y-1)+12(x-1)(y-1)$

and substitute $\displaystyle (1.01,0.99)$